What is the value of the following logarithm? $\log_{16} \left(\dfrac{1}{16}\right)$
Explanation: If $b^y = x$ , then $\log_{b} x = y$ Therefore, we want to find the value $y$ such that $16^{y} = \dfrac{1}{16}$ Any number raised to the power $-1$ is its reciprocal, so $16^{-1} = \dfrac{1}{16}$ and thus $\log_{16} \left(\dfrac{1}{16}\right) = -1$.